Structure of C 7 -Critical Graphs
5.1 No New Negative Structures are Created
In this section, we will show that no cell or vertex x with initial charge ch0(x) ≥ 0 has negative final charge after discharging. First we will show that all cells have non-negative charge at the end of the discharging process (Lemma 5.1.1). We will then prove that no vertex with degree at least four is poor after Step 5 (Lemma 5.1.2). Finally with Lemmas 5.1.3 and 5.1.4, we will prove that all vertices of degree 3 and weight at most 5 have non-negative final charge. In Section 5.2, we will prove that all vertices of degree 3 and weight at least 6 have non-negative final charge. As cells and vertices are the only structures that carry charge at any point during the discharging, this will show that the sum of the charges is non-negative, contradicting our initial assumption and completing the proof of Theorem 1.1.3.
Lemma 5.1.1. Let C be a cell in G. At the end of Step 2, ch2(C) ≥ 0.
Proof. Let X be the set of vertices in C of degree at least three. At the end of Step 1, ch1(C) =P
v∈X(15 deg(v) − 2wt(v) − 34). Rewriting,
ch1(C) = X de-pending on the degree of the cell C. Note deg(C) ≥ 3 by Lemma 2.2.3.
Case 1: deg(C) ≥ 4. Suppose that immediately after Step 1, there are p poor vertices that are each the endpoint of a short string whose other endpoint is in C. Note each of these p strings contributes at most 2 to the weight of C. For each poor vertex u we have ch1(u) ≥ −3, and so ch2(C) ≥ ch1(C) − 3p = 15 deg(C) − 28 − 2wt(C) − 3p. Since
Then X contains a single vertex of degree 3, and a single vertex of degree 4. Note all vertices that are poor immediately after Step 1 have weight at most seven, since vertices of type (4, 4, 2) are contained in cells and so send their charge to the cells that contain them in Step 1. Let v be the vertex of degree 3 in X. Let the three strings incident with v be S1, S2 and S3, named such that S1 and S2 are contained in C. Suppose that S1 is a 4-string.
By Lemma 4.1.3, S2∪ S3 is contained in a cell C0 6= C, contradicting Lemma 4.1.6. Thus we may assume S1 is not a 4-string. Symmetrically, S2 is not a 4-string.
Since |X| = 2 by assumption, we may therefore assume without loss of generality that S1 is a 3-string and S2 is a 2-string. Note S3 is therefore a 0-string. To see this, suppose not. Then by Lemma 4.1.4 applied to v and S1, we have that S2∪ S3 (and in particular, S2) is contained in either a cell or a 9-cycle C0 6= C. Suppose first C0 is a cell. Since S2 ∈ C ∩ C0, this contradicts Lemma 4.1.6 as distinct cells in G are vertex-disjoint. Thus we may assume C0 is a 9-cycle. Since S2 ∈ C ∩ C0, this contradicts Lemma 4.1.7 since cells and 9-cycles are edge-disjoint.
Thus we may assume S3 is a 0-string. Let v1 be the vertex that shares S3 with v. Note if ch2(v1) < 0, then v1 is a vertex of degree 3 not contained in a 7-cycle. By Lemma 4.1.3, v1 is thus not incident with a 4-string. Since v1 is incident with a 0-string S3, we have that wt(v1) ≤ 6. Thus if ch2(v1) < 0, it follows that ch2(v1) = −1. Let u be the vertex of degree 4 in X, and let S4 6⊆ C and S5 6⊆ C be the strings incident with u, with endpoints u1 6= u and u2 6= u, respectively. Note since ch2(C) = 17 − 2wt(C), if wt(C) ≤ 5, then ch2(C) ≥ 0 since C pays at most 1 to v1 in Step 2, and at most 3 to each of u1 and u2 in Step 2. Thus we may assume wt(C) ≥ 6. Note also if neither u1 nor u2 is poor immediately after Step 1, then ch2(C) ≥ 0 since ch2(C) ≥ 17 − 2wt(C) − ch1(v1), and wt(C) ≤ 8 by Lemma 2.2.4.
Thus we may assume at least one of u1 and u2 receives charge from C in Step 2, and so that at least one of S4 and S5 is a short string. Note that G does not contain a k-string with k ≥ 5 by Lemma 2.1.12. Furthermore, since at least one of S4 and S5 is short, S4 and S5 together contribute at most 6 to the weight of C. Thus wt(v) ≤ 6, and since wt(C) ≥ 6, it follows that wt(C) = 6. Since S3 is a 0-string and at least one of S4 and S5 is a short string, it follows that exactly one of S4 and S5 is a short string, and so that C sends charge to exactly one of u1 and u2. But then ch2(C) ≥ 0, since ch2(C) = 17 − 2wt(C) = 5 and C sends at most 1 to v1 and 3 to one of u1 and u2.
Thus we may assume |X| = 3. Since deg(C) = 3, it follows that X contains three vertices of degree three. Let S0 be a k-string with k ≤ 2 and endpoints u, v such that v ∈ V (C). Suppose that u is poor after Step 1. Then u is a degree three vertex not contained in a cell. Let S1 and S2 be the other strings incident with u.
Claim 1. The weight of u is at most six.
Proof. Suppose not. First suppose either of S1 or S2 is a 4-string. By Lemma 4.1.3 applied to u, v is contained in a cell C0 6= C, contradicting Lemma 4.1.6. Thus u is not contained in a 4-string. Suppose now k ∈ {1, 2} (i.e. that S0 is either a 1- or 2-string). First suppose that S1 is a 3-string and that S2 is not a 0-string. Then by Lemma 4.1.4 applied to u and S1, there exists an edge vv1 ∈ E(C) contained in a cell or 9-cycle C0 6= C. If C0 is a 9-cycle, this is a contradiction since 9-cycles and cells are edge-disjoint by Lemma 4.1.7.
If C0 is a cell, this too is a contradiction since distinct cells are vertex-disjoint by Lemma 4.1.6. Thus if S1 is a 3-string, we may assume S2 is a 0-string. Symmetrically, if S2 is a 3-string, we may assume S1 is a 0-string.
If k ≥ 1, then S1 and S2 therefore contribute at most 4 to the weight of C. If k = 0, since neither S1 nor S2 is a 4-string, they contribute at most 6 to the weight of C. Note by assumption k ≤ 2. Thus u has weight at most six. Since by assumption ch2(u) < 0, it follows that u has weight exactly six.
Since u has weight exactly six, we have that ch2(u) = −1.
Let A be the set of vertices that are poor immediately after Step 1, and that are incident with a short string incident with C. Let |A| − p. Note that since C has degree 3, it follows that p ≤ 3. After Step 2, we have ch2(C) = ch1(C) − p = 17 − 2wt(C) − p. Note if wt(C) ≤ 7, then since p ≤ 3 we have that ch2(C) ≥ 0, a contradiction. Thus we may assume that wt(C) ≥ 8. Note also that by Lemma 2.2.4, wt(C) ≤ 8, and so we may assume wt(C) = 8. Note also since ch2(C) < 0, we have p ≥ 2. Thus at least two of the strings incident with C each contribute at most 2 to the weight of C. Since C has weight 8, it follows that C is incident with a 4-string and so that p ≤ 2. Thus we may assume p = 2, and since wt(C) = 8, it follows that C is a (4, 2, 2)-cell.
Let S = u0u1u2u3 be a 2-string incident with C, such that u0 ∈ V (C). Note since u3
is poor after Step 2, it is not contained in a cell. Let u4 and u5 be u3’s neighbours not contained in S.
Let G0 be the graph obtained from G by identifying u4 and u5 to a new vertex z. Note G0 contains a cell of weight nine, and so G0 is not C7-critical by Lemma 2.2.4. Furthermore, G0 admits no homomorphism to C7, as any such homomorphism φ extends to G by setting φ(u4) = φ(u5) = φ(z). Therefore G0 contains a C7-critical subgraph G00, and since G006⊂ G, we have z ∈ V (G00).
Note G0 contains no 5-cycles nor triangles, since u3 ∈ V (G) is not contained in a 7-cycle by assumption. Since v(G00) ≤ v(G), G00 is not a counterexample to Theorem 1.1.3 and thus p(G00) ≤ T . Note at least one string incident with C or at least one string S0 ⊂ C is not contained in G00, as otherwise C has weight 9 in G00, contradicting Lemma 2.2.4.
Suppose first u3 6∈ V (G00). Since G00 has minimum degree at least 2 by Lemma 2.1.3, it follows that u2 6∈ V (G00). Let F be the graph obtained from G00 by splitting z back to u4, u5 and adding in the path u4u3u5. Then p(F ) = p(G00) + 17(2) − 15(2) ≤ T + 4. Since u2 6∈ V (G00) and since F ⊂ G, this contradicts Lemma 4.1.1.
We may therefore assume that u3 ∈ V (G00), and so there exists a string S0 6= S whose internal vertices are not contained in V (G00). Note either S0 is incident with C, in which case it has at least 2 internal vertices, or S0 ⊂ C, in which case V (C) ∩ V (G00) = ∅ since
Proof. Suppose not. Let A be the set of vertices that are poor immediately after Step 2
and that each share a short string with v. Let p = |A|. Note none of the p vertices in A are contained in cells, as otherwise they have charge at least 0 at the end of Step 1.
Furthermore, v is not contained in a cell, as this cell would send charge to the vertices in A in Step 2. Note since all vertices of degree 3 and weight 8 are contained in cells by Lemma 4.1.3, each vertex in A has weight at most 7, and so v sends at most 3p units of charge in Step 3.
First, suppose deg(v) ≥ 5. At the end of Step 3, we have ch3(v) ≥ ch2(v) − 3p = 15 deg(v)−2wt(v)−34−3p. Note since v is incident with p short strings and the remaining deg(v) − p strings incident with v each contribute at most 4 to the weight of v, it follows that wt(v) ≤ 4(deg(v) − p) + 2p. Thus ch3(v) ≥ 15 deg(v) − 2(4 deg(v) − 2p) − 34 − 3p = 7 deg(v) − 34 + p ≥ 1 + p, since deg(v) ≥ 5. Since p is non-negative, ch3(v) ≥ 0, a contradiction.
Thus we may assume deg(v) = 4. Since v is not contained in a cell, by Lemma 4.1.5 v is not incident with a 4-string. Since v is incident with at least p short strings and at most (deg(v) − p) k-string with k = 3, it follows that v has weight at most 3(deg(v) − p) + 2p = 12 − p, and so
ch3(v) ≥ ch2(v) − 3p = 15 deg(v) − 2wt(v) − 34 − 3p
≥ 60 − 2(12 − p) − 34 − 3p
= 2 − p.
Thus if p ≤ 2, we have ch3(v) ≥ 0, a contradiction.
Since p ≤ deg(v) = 4, we may therefore assume p ∈ {3, 4}. Note if wt(v) ≤ 7, since ch3(v) ≥ 15 deg(v) − 2wt(v) − 34 − 3p
≥ 60 − 14 − 34 − 3p
= 12 − 3p it follows that ch3(v) ≥ 0, a contradiction.
Thus v has weight at least 8 and is incident with at least three short strings. It follows that v is either a vertex of type (3, 2, 2, 2), (2, 2, 2, 2), or (3, 2, 2, 1). Note if v shares a short string with a poor vertex v0 of weight 6, then v pays v0 only 1 and each of the other (p − 1) vertices in A at most 3. Thus v pays the vertices in A at most 3(p − 1) + 1, and so it follows that ch3(v) ≥ 4 − p. Since p ∈ {3, 4}, this is non-negative. Thus we may assume that v pays at least three poor vertices of weight 7. We finish the proof with the following three claims.
Claim 1. v is not a vertex of type (2, 2, 2, 2).
Proof. Suppose not. Recall v only pays a subset of the vertices of degree 3 that are not contained in cells. Furthermore, since v is of type (2, 2, 2, 2), v only pays vertices that are incident with 2-strings. Since the only vertices of degree 3 and weight 7 not contained in cells and incident with 2-strings are vertices of type (3, 2, 2), we may assume v pays at least three vertices of type (3, 2, 2). Since each vertex of type (3, 2, 2) in A is not in a cell, by Lemma 4.1.4 they are each contained in 9-cycles. Thus there exist vertices a 6= b in A of type (3, 3, 2), such that Sa = aa1a2v is a 2-string shared by v and a, Sb is a 2-string shared by v and b, and Sab is a 2-string shared by a and b. Similarly, there exist vertices c 6= d in A such that c is of type (3, 2, 2), Sc is a 2-string shared by v and c, Sdis a 2-string shared by v and d, and Scd is a 2-string shared by c and d. Let C be the 9-cycle formed by Sa∪ Sb ∪ Sab. Let G0 be the graph obtained from G by identifying v and a1 to a new vertex z, and deleting a2. Note G0 does not admit a homomorphism φ to C7 as otherwise φ extends to G by setting φ(v) = φ(a1) = φ(z) and φ(a2) ∈ NC7(φ(z)). Furthermore, G0 itself is not C7-critical, as the cycle C0 obtained from C with the identification of v and a1 is a cell of weight 10 in G0, contradicting Lemma 2.2.4. Thus G0 contains a C7-critical subgraph G00.
Note G00 does not contain at least one string incident with C0 or one string in E(C0).
We claim one of these strings not contained in G00 contains at least one internal vertex.
To see this, suppose not. Note C0 is incident with four strings: Sc, Sd, and two 3-strings Sa0 and Sb0 incident with a and b, respectively. Since these four strings all have internal vertices, we may assume they are all contained in G00. Since Sab and Sb are both 2-strings, we may assume they are both contained in G00. Thus G0 does not contain the edge az. But then Sab∪ Sa0 is a 6-string contained in G00, contradicting Lemma 2.1.12.
Thus we may assume there is a vertex in V (G) \ {a1, a2, v} that is not contained in V (G00).
Since G00 6⊂ G, it follows that G00 contains the new vertex z. Note G00 does not contain a triangle or 5-cycle, since Sa is contained in a 9-cycle in G and so is not contained in a cell by Lemma 4.1.7. Since v(G00) < v(G0), it follows that p(G00) ≤ T .
Let F be the graph obtained from G00by splitting z back into the vertices v and a1 and adding in the path va2a1. Then p(F ) = p(G00) + 17(2) − 15(2) ≤ T + 4. But since F 6= G, this contradicts Lemma 4.1.1.
Claim 2. v is not a vertex of type (3, 2, 2, 2).
Proof. Suppose not. Since v is of type (3, 2, 2, 2), v only pays vertices that are incident with 2-strings. Since the only vertices of degree 3 and weight 7 not contained in cells and incident with 2-strings are vertices of type (3, 2, 2), we may assume v pays three vertices
of type (3, 2, 2). Since each vertex of type (3, 2, 2) in A is not in a cell, by Lemma 4.1.4 they are each contained in 9-cycle. Let a 6= b 6= c be the three vertices of type (3, 2, 2) in A. Let Sa, Sb, and Sc be the 2-strings shared by v with a, b, and c, respectively. Let Sa0 be the other 2-string incident with a. By Lemma 4.1.4, Sa0 ∪ Sa is contained in a 9-cycle C. Since v has degree exactly 4 and is incident only with 2-strings and a 3-string, one of Sb and Sc is contained in C. Without loss of generality, we may assume C = Sa∪ Sa0 ∪ Sb, and so that Sa0 is shared by b and a. Similarly, let Sc0 6= Sc be a 2-string incident with c. By Lemma 4.1.4, Sc∪ Sc0 is contained in a 9-cycle C0 6= C. Since v has degree exactly 4 and is incident only with 2-strings and a 3-string, it follows that there exists a 2-string S 6= Sc incident with v that is contained in C0. Thus Sa or Sb is contained in C0. But this contradicts Lemma 4.1.8, since Sb∪ Sa ⊂ C.
The only remaining possibility is then that v is of type (3,2,2,1). Since |A| ≥ 3 and each vertex in A has weight 7, we may assume v shares its incident 2-strings S1 and S2 with two poor (3, 2, 2) vertices u1 and u2, and its incident 1-string S3 with a poor (3, 3, 1) vertex u3. By applying Lemma 4.1.4 to u1, u2, and u3, we find each of S1 and S2 is contained in two 9-cycles, contradicting Lemma 4.1.8.
Lemma 5.1.3. Let v ∈ V (G) be a vertex of degree 3 and weight at most 4. At the end of Step 4, ch4(v) ≥ 0.
Proof. Suppose not. Let A be the set of vertices that are poor immediately after Step 4 and that each share a short string with v. Let p = |A|. Note none of the p vertices in A are contained in cells, as otherwise they have charge at least 0 at the end of Step 1.
Furthermore, v is not contained in a cell, as otherwise this cell sends charge to the vertices in A in Step 2. Note since all vertices of degree 3 and weight 8 are contained in cells by Lemma 4.1.3, each vertex in A has weight at most 7, and so v sends at most 3p units of charge in Step 3.
Note if A contains p ≤ 3 vertices of weight 6, then v sends only 3 units of charge in Step 4, and so
ch4(v) ≥ ch3(v) − 3
= 15 deg(v) − 2wt(v) − 34 − 3
≥ 45 − 8 − 34 − 3
= 0, a contradiction.
We may therefore assume A contains a vertex u of weight at least 7. Since A only con-tains vertices of weight at most seven, we may assume u has weight exactly 7. Furthermore,
if p = 1 (and so v sends charge to only one vertex in Step 4), then ch4(v) = ch3(v) − 3 = 0, a contradiction. Thus we may assume p ≥ 2. Since vertices of type (4, 3, 0) and (4, 2, 1) are contained in cells and thus have charge 0 after Step 1, we may assume u is either of type (3, 3, 1) or of type (3, 2, 2).
First suppose u is of type (3, 2, 2). Let u1 be the other vertex that shares a 2-string with u. Note if u1 is contained in a cell, then ch4(u) = 0 by Step 2, and so u 6∈ A, a contradiction. Furthermore, if deg(u1) ≥ 4, then ch4(u) = 0 by Step 3, and so again u 6∈ A. We may therefore assume u1 has degree three. But then this contradicts Lemma 4.2.5, as neither v nor u1 is contained in a cell.
Thus we may assume u is of type (3, 3, 1). Let S1 and S3 be the two 3-strings incident with u, and let S2 be the 1-string shared by u and v (see Figure 5.1. By Lemma 4.1.4, since u is a (3, 3, 1)-vertex not contained in a cell, we have that S2∪ S3 is contained in a 9-cycle C. Similarly, S1∪ S2 is contained in a 9-cycle C0 6= C. Since both 9-cycles contain S2, we have that C0∩ C = S2 by Lemma 4.1.8. Let S4 and S5 be the other two strings incident with v. Without loss of generality, suppose S4 ⊂ C0 and S5 ⊂ C. Note since p ≥ 2, one of S4 and S5 is shared by v with a vertex in A. Without loss of generality, we may assume S4 is shared by v and a vertex w in A.
Since v has weight at most four and is incident with a 1-string S2, it follows that at least one of S4 and S5 is not a 2-string.
First suppose S4 is not a 2-string. Since w ∈ A, we have that w has degree three and wt(w) ≥ 6. Furthermore, since w is not contained in a cell, w is not incident with a 4-string by Lemma 4.1.3. Since the internal vertices of two of the strings incident with w are contained in C0 \ V (S1 ∪ S2), together these two strings contribute at most 1 to the weight of w. But then w has weight at most 4, a contradiction.
Thus we may assume S4 is a 2-string, and since w has weight at least 6, we have that w is a vertex of type (3, 2, k) with k ≥ 1.
Let S6 denote the third string incident with w (so w is incident with S1, S4, and S6). By Lemma 4.1.4 applied to w, S4 and S6 are contained in a 9-cycle C00. Since S4 is contained in C0∩ C00, this contradicts Lemma 4.1.8.
Lemma 5.1.4. Let v ∈ V (G) be a vertex of degree 3 and weight 5 that shares a short string with only one poor vertex at the end of Step 4. At the end of Step 5, ch5(v) ≥ 0.
Proof. Suppose not. Let u be the vertex with ch4(u) < 0 that shares a short string with
v u S1
S2
S3
S4 S5
Figure 5.1: Figure for Lemma 5.1.3. v has weight at most four and degree three. The white vertices are of unknown degree, though their degree is at least that shown. The black vertices’ degrees are as illustrated.
v. Suppose wt(u) = 6. Then ch4(u) = −1, and so ch5(v) = ch4(v) − ch4(u)
= 15 deg(v) − 2wt(v) − 34 − 1
= 45 − 10 − 34 − 1
= 0, a contradiction.
We may therefore assume u has weight at least seven.
Note since ch4(u) < 0, u is not contained in a cell by the discharging rules. By Lemma 4.1.3, it follows that u is not of type (4,2,2), (4,3,0), or (4,2,1). Thus we may assume u is a vertex of type either (3,3,1), or (3,2,2). By Lemma 4.2.5, if u is a (3,2,2)-vertex it has charge at least 0 by either Step 1, 2, or 3.
Therefore we may assume u is a vertex of type (3,3,1), and hence the string shared by u and v is a 1-string. Note v is not incident with a 4-string as otherwise it is contained in a cell by Lemma 4.1.5 and so ch2(u) ≥ 0, contradicting the fact that u ∈ A. Thus since v has degree 3 and weight 5 and is incident with a 1-string, it is either a vertex of type (3,1,1) or of type (2,2,1).
First suppose v is of type (3, 1, 1). Let S1 and S2 be the two 1-strings incident with v, named such that S1 is shared by u and v. Let S3 be the 3-string incident with v. Let the two 3-strings incident with u be named S4 and S5. Note S4 and S5 do not have the same two endpoints by Lemma 2.2.6. Furthermore v and u share only S1 by Lemma 2.2.6. By Lemma 4.1.4 applied to u and S4, since u is not contained in a cell S1∪ S5 is contained in a 9-cycle C. Note since v has degree 3, it follows that either S2 or S3 is contained in C.
Since v(S1∪ S5∪ S3) = 11, we may assume that S2 ⊂ C.
Similarly, by Lemma 4.1.4 applied to u and S5, we have that S4 ∪ S1 is contained in a 9-cycle C0. Note since v has degree 3, it follows that either S2 or S3 is contained in C0. Since v(S1∪ S4∪ S3) = 11, it follows that S2 ⊂ C0.
Since S1 ∪ S2 ⊂ C ∩ C0, this contradicts Lemma 4.1.8.
Therefore we may assume v is a (2, 2, 1)-vertex. Let a and b be the two vertices that share a 2-string with v. Note a 6= b by Lemma 2.2.6. Let Sa be the string shared by a and v, and let Sb be the 2-string shared by b and v. Let S1, S2 and S3 be the three strings incident with u, named such that S1 is shared by u and v. Note S2 and S3 do not have the same endpoints by Lemma 2.2.6. By Lemma 4.1.4 applied to u and S2, since u is not contained in a cell we have that S3∪ S1 is contained in a 9-cycle C. Similarly, by applying Lemma 4.1.4 to u and S3, we have that S2∪ S1 is contained in a 9-cycle C0. Thus without loss of generality, we may assume S2 is shared by a and u, and that S3 is shared by b and u.
Let G0 be the graph obtained from G by deleting u, v, and all of the internal vertices of their incident strings. Since G is C7-critical, G0 has a homomorphism φ to a cycle C = c1c2c3c4c5c6c7c1. Without loss of generality, we may assume φ(a) = c1 and φ(b) ∈ {c1, c2, c3, c4}. Since φ does not extend to G, φ(b) = c4. (To see the extensions of all other homomorphisms to G, see Figure 5.2.) Let G00 be the graph obtained from G0 by adding a new vertex z and edges az and bz. Note now the following:
Claim 1. There does not exist a homomorphism φ : G00 → C with φ(a) = c1 and φ(b) = c4.
Proof. Suppose φ : G00 → C is such that φ(a) = c1. Note φ(b) ∈ Bφ(b|a, azb). But Bφ(b|a, azb) = NC(NC(c1)) = {c1, c3, c6}.
Thus if G00 admits a homomorphism to C, this homomorphism extends to a homomor-phism of G to C, since by Claim 1 there does not exist a homomorhomomor-phism φ from G00 to C with φ(a) = c1 and φ(b) = c4. We may thus assume G00 contains a C7-critical subgraph G000, and since G000 6⊂ G, we have z ∈ V (G000). Furthermore, since G000 has minimum degree at least two, {az, zb} ∈ E(G000).
Suppose G000 is a triangle. Then ab is an edge in E(G). But then abSaSb is a cell C00 with Sa ⊂ C00∩ C0, contradicting Lemma 4.1.7. Suppose now G000 is a 5-cycle. Then there exists an (a, b)-path P of length 3. But then P ∪ Sa∪ Sb is a 9-cycle C00 with Sa⊂ C00∩ C0, contradicting Lemma 4.1.8.
Thus we may assume that G000 is not a triangle or 5-cycle. Since v(G000) < v(G), it follows that G000 is not a counterexample to Theorem 1.1.3, and so p(G000) ≤ T . Let F be the graph
c4
Figure 5.2: Figure for Claim 5.1.4. Extensions of φ to G. The white vertices are of unknown degree, though their degree is at least that shown. The black vertices’ degrees are as illustrated.
obtained from G000 by deleting z and adding Sa∪ Sb. Then p(F ) = p(G000) + 17(4) − 15(4) ≤
obtained from G000 by deleting z and adding Sa∪ Sb. Then p(F ) = p(G000) + 17(4) − 15(4) ≤